To solve the equation (4x-3)(4x+3)-(6x-1)2=0, we will first simplify the expression on the left side.

Expanding the first term: (4x-3)(4x+3)
= 16x^2 + 12x - 12x - 9
= 16x^2 - 9

Expanding the second term: (6x-1)2
= 12x - 2

Now the equation becomes:
16x^2 - 9 - 12x + 2 = 0
16x^2 - 12x - 7 = 0

Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 16, b = -12, and c = -7.

To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a

Plugging in the values:
x = (12 ± sqrt((-12)^2 - 416(-7))) / 2*16
x = (12 ± sqrt(144 + 448)) / 32
x = (12 ± sqrt(592)) / 32
x = (12 ± 24.33) / 32

Now we have two possible solutions:
x = (12 + 24.33) / 32 = 36.33 / 32 ≈ 1.135
x = (12 - 24.33) / 32 = -12.33 / 32 ≈ -0.385

Therefore, the solutions to the equation are x ≈ 1.135 and x ≈ -0.385.